The figure is not drawn to scale. CEFG is a parallelogram, BCG and CDG are isosceles triangles. GF//HR and ∠FPR= 105°. BE is straight line.
- Find ∠BGC.
- Find ∠EDG.
(a)
∠FPR = 105°
∠GLN = 105° (Corresponding angles)
∠KLG
= 180° - 105°
= 75° (Angles on a straight line)
∠BGC
= 180° - 75° - 75°
= 30° (Isosceles triangle)
(b)
∠KLG = ∠BCG = 75° (Corresponding angles)
∠GCD
= 180° - 75°
= 105° (Angles on a straight line)
∠CDN
= (180° - 105°) ÷ 2
= 37.5° (Isosceles triangle)
∠EDG
= 180° - 37.5°
= 142.5° (Angles on a straight line)
Answer(s): (a) 30°; (b) 142.5°